Question 11.CA.7: A block of weight W is dropped from a height h onto the free......
A block of weight W is dropped from a height h onto the free end of the cantilever beam AB (Fig. 11.23). Determine the maximum value of the stress in the beam.
As it falls through the distance h, the potential energy Wh of the block is transformed into kinetic energy. As a result of the impact, the kinetic energy is transformed into strain energy. Therefore,
U_m=W h (1)
The total distance the block drops is actually h+y_m, \text { where } y_m is the maximum deflection of the end of the beam. Thus a more accurate expression for U_m (see Sample Prob. 11.3) is
U_m=W\left(h+y_m\right) (2)
However, when h \gg y_m, y_m may be neglected, and thus Eq. (1) applies.
Recalling the equation for the strain energy of the cantilever beam AB in Concept Application 11.3, which was based on neglecting the effect of shear,
U_m=\frac{P_m^2 L^3}{6 E I}
Solving this equation for P_m , the static force that produces the same strain energy in the beam is
P_m=\sqrt{\frac{6 U_m E I}{L^3}} (3)
The maximum stress σ_m occurs at the fixed end B and is
\sigma_m=\frac{|M| c}{I}=\frac{P_m L c}{I}
Substituting for P_m from Eq. (3),
\sigma_m=\sqrt{\frac{6 U_m E}{L\left(I / c^2\right)}} (4)
or recalling Eq. (1),
\sigma_m=\sqrt{\frac{6 W h E}{L\left(I / c^2\right)}}